# Is $\phi B(\omega,\;\cdot\;)$ Lebesgue integrable over $[0,\infty)$ for a real-valued Brownian motion $B$ and $\phi\in C_c^\infty(\mathbb R)$?

Let $(B_t)_{t\ge 0}$ be a real-valued Brownian motion on a probability space $(\Omega,\mathcal A,\operatorname P)$ and $\lambda$ be the Lebesgue measure on $\mathbb R$. Is $$W(\phi):=\int_{[0,\infty)}\phi(t)B_t\;{\rm d}\lambda\;\;\;\text{for }\phi\in C_c^\infty(\mathbb R)$$ well-defined?

Let $\phi\in C_c^\infty(\mathbb R)$ and $K:=\operatorname{supp}\phi\cap [0,\infty)$. I would argue that $$\left\|\phi1_{[0,\infty)}B(\omega,\;\cdot\;)\right\|_{L^1(\lambda)}\le\underbrace{\left\|\phi\right\|_{L^\infty(\lambda)}}_{<\infty}\underbrace{\left\|1_KB(\omega,\;\cdot\;)\right\|_{L^1(\lambda)}}_{=:M(\omega)}\;\;\;\text{for all }\omega\in\Omega$$ and that $M<\infty$ almost surely, since $K$ is compact and $B$ is almost surely continuous.

How do I deal with the null set on which $B$ is not continuous? Please note that $W$ is the distributional stochastic process associated with $B$. Maybe we don't care about its values on a null set.

## 1 Answer

$M<\infty$ because $B$ is almost surely continuous and thus bounded when restricted to a compact set

• That's what I've said, but we only have $M<\infty$ almost surely. – 0xbadf00d Jan 30 '16 at 12:39
• @0xbadf00d Almost surely is the most you can possibly say about any property of $B_t$. – David C. Ullrich Jan 30 '16 at 13:12
• It is good practice to make edits to your question visible as such, especially when those edits are motivated by a response to your question. What you said in the question I answered to was that M was finite because of Riemann integrability of the Brownian motion, and you asked whether this argument was correct. And David is right, when you give me a Brownian motion, I can change it on a null set to be constant infinity and it will still be a Brownian motion – Bananach Jan 30 '16 at 17:50
• @Bananach But it's essentially the same argument. Since $K$ is compact and $B$ is continuous, $B$ is Riemann integrable and the Riemann integral and the Lebesgue integral coincide. I've edited the question cause I want to make clear that my real question is how I need to handle the null set. – 0xbadf00d Jan 30 '16 at 23:57
• It is essentially the same, indeed. But with your current edit it is the exact same. Therefore it would be nice to format your question in a way to reflect the change of focus. But no hard feelings, the null set question is clear now? – Bananach Jan 31 '16 at 8:05